Giải phương trình
$$(1+\sqrt{3})sin \left (2x + \frac{\pi}{4}\right )=2\sqrt{2} \left [cos \left (x-\frac{\pi}{3}\right )-sin^{2}x \right ]$$
Toán thủ ra đề
nguyenhang28091996
Lordsky 216 xin giải đề :
$$PT\Leftrightarrow (1+\sqrt{3})(sin2x + cos 2x)=4 [\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx -\frac{1-cos2x}{2}]$$
$$\Leftrightarrow (1+\sqrt{3})sin2x+(1+\sqrt{3})cos2x=2cosx+2\sqrt{3}sinx -2 +2cos2x$$
$$\Leftrightarrow (1+\sqrt{3})sin2x+(\sqrt{3}-1)cos2x=2cosx+2\sqrt{3}sinx -2$$
$$\Leftrightarrow (2-cos2x+\sqrt{3}sin2x)+sin2x+\sqrt{3}cos2x=2(cosx+\sqrt{3}sinx)$$
$$\Leftrightarrow (3sin^2x+2\sqrt{3}sinxcosx+cos^2x)+ 2 sinx cosx+\sqrt{3}cos^2x-\sqrt{3}sin^2x=2(cosx +\sqrt{3}sinx)$$
$$\Leftrightarrow (\sqrt{3}sinx+cosx)^2-2(\sqrt{3}sinx+cosx)+\sqrt{3}cos^2x-3\sqrt{3}sin^2x+2sinx(\sqrt{3}sinx+cosx)=0$$
$$\Leftrightarrow (\sqrt{3}sinx+cosx)^2-2(\sqrt{3}sinx+cosx)+\sqrt{3}(cosx-\sqrt{3}sinx)(\sqrt{3}sinx+cosx)+2sinx(\sqrt{3}sinx+cosx)=0$$
$$\Leftrightarrow (\sqrt{3}sinx+cosx)(\sqrt{3}sinx+cosx-2+\sqrt{3}cosx-3sinx+2sinx)=0$$
$$\begin{bmatrix}
\sqrt{3}sinx+cosx=0 (1) & \\ \sqrt{3}sinx-sinx+cosx+\sqrt{3}cosx-2=0(2)
&
\end{bmatrix}$$
Xét
$$(1)\Leftrightarrow sin(x+\frac{\Pi}{6} )=0
\Leftrightarrow x+\frac{\Pi}{6}=k\Pi(k \in \mathbb{Z})\Leftrightarrow x=\frac{-\Pi}{6}+k\Pi$$
Xét
$$(2)\Leftrightarrow (\sqrt{3}-1)sinx+(\sqrt{3}+1)cosx=2$$
$$\Leftrightarrow \frac{\sqrt{3}-1}{2\sqrt{2}}sinx+\frac{\sqrt{3}+1}{2\sqrt{2}}cosx=\frac{\sqrt{2}}{2}$$
GỌi $cos\alpha =\frac{\sqrt{3}-1}{2\sqrt{2}}$, $ sin\alpha =\frac{\sqrt{3}+1}{2\sqrt{2}}$
CÓ:
$$(2)\Leftrightarrow sinx.cos\alpha +cosx.sin\alpha =\frac{\sqrt{2}}2{}$$
$$\Leftrightarrow sin(x+\alpha )=\frac{\sqrt{2}}{2}$$
$$ \Leftrightarrow \begin{bmatrix}
x+\alpha =\frac{\Pi}{4}+k2\Pi & \\ x+\alpha=\frac{3\Pi}{4} +k2\Pi
&
\end{bmatrix}$$
Vậy $\begin{bmatrix}
x=\frac{-\pi}{6}+k\pi & \\ x=-\alpha +\frac{3\Pi}{4}+k2\Pi
& \\ x=-\alpha +\frac{\pi}{4}+k2\pi
&
\end{bmatrix}$
$$\boxed{\boxed{Điểm: 9}}$$
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- L Lawliet yêu thích